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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Mar 14, 2016 at 22:18 comment added Salvo Tringali Sorry for the delay, I was in a hurry this afternoon and didn't realize that there were at least three things I had to fix in the old formulation. In particular, I would/should have asked if $\mathbb B$ can be embedded into $\mathbb A$ in such a way that at least one $x\in X$ becomes right-subtractive. @BenjaminSteinberg. Yes, and if $\mathbb A$ is a canc. monoid, then the right-subtractive elements of a finite set $X$ that contains the identity of $\mathbb A$ are precisely the left-invertible elements of $\mathbb A$ (hence $X^{\rm rs}=\mathbb A^\times$, from what you made me note yesterday).
Mar 14, 2016 at 21:59 history edited Salvo Tringali CC BY-SA 3.0
Changed a word in the title
Mar 14, 2016 at 21:49 history edited Salvo Tringali CC BY-SA 3.0
Fixed a couple of issues in the formulation of the question
Mar 14, 2016 at 17:47 comment added Benjamin Steinberg If $X$ is just the identity, then isn't $X$-subtractive the same as left-invertible?
Mar 14, 2016 at 14:40 history edited Salvo Tringali CC BY-SA 3.0
added 10 characters in body
Mar 14, 2016 at 14:34 history asked Salvo Tringali CC BY-SA 3.0